Antique Car Auction

A man is going to an Antique Car auction. All purchases must be paid for in cash. He goes to the bank and draws out $25,000.

Since the man does not want to be seen carrying that much money, he places it in 15 envelopes numbered 1 through 15. Each envelope contains the least number of bills possible of any available US currency (for example, no two tens instead of a twenty).

At the auction he makes a successful bid of $8322 for a car. He hands the auctioneer envelopes 2, 8, and 14. After opening the envelopes the auctioneer finds exactly the right amount.

How many ones did the auctioneer find in the envelopes?

 

Answer:

One.
Unfortunately, Lucas Jones didn't provide a more elaborate explanation. But thanks to Ronald A. Laski we can now present a more acceptable explanation:

In Envelope #14, there are:
    81 $100 bills = $8100
     1 $ 50 bill  = $  50
     2 $ 20 bills = $  40
     1 $  2 bill  = $   2
                    ----- +
                    $8192 (total amount of money for Envelope #14) In Envelope #8, there are:      1 $100 bill  = $100
     1 $ 20 bill  = $ 20
     1 $  5 bill  = $  5
     1 $  2 bill  = $  2
     1 $  1 bill  = $  1 <- that's the one!
                    ----
                    $128 (total amount of money for Envelope #8)

Envelope #2 has 1 $2 bill in it, which is its total amount also.
Now $8192 + $128 + $2 = $8322. And that is the winning bid!
The amount of money in each envelope is 2 ^ (# of envelope - 1), except for envelope 15, which contains $8617.